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This question seems obvious and yet I can't seem to find a good answer anywhere.

Let $V$ be a finite dimensional vector space, and let $0$ denote the trivial vector space. Is $V \otimes 0 = 0$ or $V \otimes 0 = V$? My gut tells me that it is the second case, but in thinking about dimension, tensor product should multiply dimension in which case I think it is the first case.

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    $\begingroup$ The elements of $V\otimes 0$ are sums of pure tensors of the form ${\bf v}\otimes{\bf 0}$. Then, as we can slide scalars across the $\otimes$ symbol, we have ${\bf v}\otimes{\bf 0}={\bf v}\otimes 0{\bf 0}=0{\bf v}\otimes{\bf 0}={\bf 0}\otimes{\bf 0}$. Thus, there is only one element in $V\otimes 0$, the zero tensor. $\endgroup$ – runway44 Jul 24 at 22:54
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    $\begingroup$ @runway44 that is IMO a better answer than the two given ones. Why did you post it as a comment? $\endgroup$ – leftaroundabout Jul 25 at 10:11
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Notice that the space of bilinear maps $f:V\times \{0\}\to k$ consists of exactly the zero map, therefore the constant map $w:V\times \{0\}\to\{0\}$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:V\times\{0\}\to k$ is the constant zero map, therefore the linear map $0:\{0\}\to k$ satisfies $f=0\circ w$. $0$ is also the only linear map $\{0\}\to k$, so it's a fortiori the only linear map $g$ such that $f=g\circ w$.

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Recall that the tensor product $V\otimes W$ of two finite-dimensional vector spaces $V$ and $W$ satisfy the dimension formula $\dim(V\otimes W)=\dim(V)\cdot\dim(W)$.

So, tensoring a finite-dimensional vector space $V$ with the trivial vector space $0$ yields a vector space $V\otimes 0$ with dimension $$ \dim(V\otimes 0)=\dim(V)\cdot\dim(0)=\dim(V)\cdot 0 = 0 $$ This implies that $V\otimes0$ is itself the trivial vector space $V\otimes 0=0$.

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The elements of $V\otimes 0$ are sums of pure tensors of the form ${\bf v}⊗{\bf 0}$. Then, as we can slide scalars across the $\otimes$ symbol, we have ${\bf v}\otimes{\bf 0}={\bf v}\otimes (0\cdot{\bf 0})=(0\cdot{\bf v})\otimes{\bf 0}={\bf 0}\otimes{\bf 0}$.

Thus, there is only one element in $V\otimes 0$, the zero tensor.

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